Categories: Mathematics, Physics.

# Fundamental solution

Given a linear operator $\hat{L}$ acting on $x \in [a, b]$, its fundamental solution $G(x, x')$ is defined as the response of $\hat{L}$ to a Dirac delta function $\delta(x - x')$ for $x \in ]a, b[$:

\begin{aligned} \boxed{ \hat{L}\{ G(x, x') \} = A \delta(x - x') } \end{aligned}

Where $A$ is a constant, usually $1$. Fundamental solutions are often called Green’s functions, but are distinct from the (somewhat related) Green’s functions in many-body quantum theory.

Note that the definition of $G(x, x')$ generalizes that of the impulse response. And likewise, due to the superposition principle, once $G$ is known, $\hat{L}$’s response $u(x)$ to any forcing function $f(x)$ can easily be found as follows:

\begin{aligned} \hat{L} \{ u(x) \} = f(x) \quad \implies \quad \boxed{ u(x) = \frac{1}{A} \int_a^b f(x') \: G(x, x') \dd{x'} } \end{aligned}

$\hat{L}$ only acts on $x$, so $x' \in ]a, b[$ is simply a parameter, meaning we are free to multiply the definition of $G$ by the constant $f(x')$ on both sides, and exploit $\hat{L}$’s linearity:

\begin{aligned} A f(x') \: \delta(x - x') = f(x') \hat{L}\{ G(x, x') \} = \hat{L}\{ f(x') \: G(x, x') \} \end{aligned}

We then integrate both sides over $x'$ in the interval $[a, b]$, allowing us to consume $\delta(x \!-\! x')$. Note that $\int \dd{x'}$ commutes with $\hat{L}$ acting on $x$:

\begin{aligned} A \int_a^b f(x') \: \delta(x - x') \dd{x'} &= \int_a^b \hat{L}\{ f(x') \: G(x, x') \} \dd{x'} \\ A f(x) &= \hat{L} \int_a^b f(x') \: G(x, x') \dd{x'} \end{aligned}

By definition, $\hat{L}$’s response $u(x)$ to $f(x)$ satisfies $\hat{L}\{ u(x) \} = f(x)$, recognizable here.

While the impulse response is typically used for initial value problems, the fundamental solution $G$ is used for boundary value problems. Suppose those boundary conditions are homogeneous, i.e. $u(x)$ or one of its derivatives is zero at the boundaries. Then:

\begin{aligned} 0 &= u(a) = \frac{1}{A} \int_a^b f(x') \: G(a, x') \dd{x'} \qquad \implies \quad G(a, x') = 0 \\ 0 &= u_x(a) = \frac{1}{A} \int_a^b f(x') \: G_x(a, x') \dd{x'} \quad \implies \quad G_x(a, x') = 0 \end{aligned}

This holds for all $x'$, and analogously for the other boundary $x = b$. In other words, the boundary conditions are built into $G$.

What if the boundary conditions are inhomogeneous? No problem: thanks to the linearity of $\hat{L}$, those conditions can be given to the homogeneous solution $u_h(x)$, where $\hat{L}\{ u_h(x) \} = 0$, such that the inhomogeneous solution $u_i(x) = u(x) - u_h(x)$ has homogeneous boundaries again, so we can use $G$ as usual to find $u_i(x)$, and then just add $u_h(x)$.

If $\hat{L}$ is self-adjoint (see e.g. Sturm-Liouville theory), then the fundamental solution $G(x, x')$ has the following reciprocity boundary condition:

\begin{aligned} \boxed{ G(x, x') = G^*(x', x) } \end{aligned}

Consider two parameters $x_1'$ and $x_2'$. The self-adjointness of $\hat{L}$ means that:

\begin{aligned} \int_a^b G^*(x, x_1') \Big( \hat{L} \{ G(x, x_2') \} \Big) \dd{x} &= \int_a^b \Big( \hat{L} \{ G(x, x_1') \} \Big)^* G(x, x_2') \dd{x} \\ \int_a^b G^*(x, x_1') \: \delta(x - x_2') \dd{x} &= \int_a^b \delta^*(x - x_1') \: G(x, x_2') \dd{x} \\ G^*(x_2', x_1') &= G(x_1', x_2') \end{aligned}
1. O. Bang, Applied mathematics for physicists: lecture notes, 2019, unpublished.